Possible values of $\gcd(2a^2+6a-4, 2a^2+4a-3)$

Question

Let $a \in \mathbb{Z}$.Calculate all possible values of $\gcd(2a^2+6a-4, 2a^2+4a-3)$.

By factoring, I can show that they have no common factors, so the answer is 1, but I have to go through non-integers. How can we get to the same result without stepping outside of $\mathbb{Z}$?

Answer 1

Any common factor of $2a^2+6a-4$ and $2a^2+4a-3$ is also a factor of

• $(2a^2+6a-4)-(2a^2+4a-3)=2a-1$
• and hence of $(2a^2+4a-3)-a(2a-1)=5a-3$
• and hence of $5(2a-1)-2(5a-3)=1$.

If you wish you can combine all these to get a more direct but much less obvious solution using the fact that $$(2a+5)(2a^2+6a-4)-(2a+7)(2a^2+4a-3)=1\ .$$